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10 questions
3
votes
1
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392
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A combinatorial identity involving binomial coefficients
When I was reading an article by CHUN-GANG JI (A SIMPLE PROOF OF A CURIOUS CONGRUENCE BY ZHAO), he mentioned in the acknowledgement
the following identity
$$\sum_{i+j+k=p,\text{ } i,j,k\gt 0}{p\choose ...
- 53
3
votes
0
answers
274
views
Inequalities for Motzkin polynomials
Let us denote by $M_{n}(t)$ the $n$-th Motzkin polynomial. It is defined by $M_1(t) = M_2(t) = 1$ and
$$ M_{n}(t) = \sum_{i=0}^{\lfloor n/2\rfloor } \frac{1}{n-1-i} \binom{n-1-i}{i} \binom{n-1}{i+1} t^...
- 1,889
4
votes
1
answer
376
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Counting permutations with a fixed number of descents and an extra condition
I am computing the volumes of certain polytopes and it turns out that knowing a "closed formula" for the following number would help a lot.
Determine the number of permutations $\sigma\in \...
- 1,889
2
votes
3
answers
742
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Asking for a proof for a sum of products of binomials: an "interesting" identity?
The following identity must have received alternative proofs, including a combinatorial argument by David Callan as found at Bijections for the Identity $4^n = \sum_{k = 0}^n \binom{2k}k\binom{2(n - k)...
- 43.1k
3
votes
3
answers
396
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Chebyshev polynomials and ballot numbers
I have asked this question a short time ago on mathstackexchange, but it has already fallen into the abyss of answered and uncommented questions. So I take the risk to ask it on mathoverflow.
Playing ...
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11
votes
3
answers
557
views
In search of a $q$-analogue of a Catalan identity
Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\...
- 43.1k
12
votes
1
answer
730
views
Two remarkable weighted sums over binary words
This question builds off of the previous MO question Number of collinear ways to fill a grid.
Let $A(m,n)$ denote the set of binary words $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_{m+n-2})$ consisting ...
- 24.2k
12
votes
3
answers
1k
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A "quantum" identity: in search of a proof -Part II
As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on this ...
- 43.1k
12
votes
2
answers
1k
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An interesting identity: in search of a proof -Part I
I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS.
Question. Can you show that
$$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-...
- 43.1k
3
votes
1
answer
253
views
What is the value of this sum involving q-binomials?
Let $n\ge 2r$ be positive integers.
Is there a closed form for following finite summation involving in q-binomial coefficients
$$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
- 1,093